Question

The space shuttle launches an 850-kg satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.8 s to give it a velocity of\(0.30\;{\rm{m/s}}\)in the x direction relative to the shuttle. The mass of the shuttle is 92,000 kg. (a) Determine the component of velocity\({v_{\rm{f}}}\)of the shuttle in the minus x direction resulting from the ejection. (b) Find the average force that the shuttle exerts on the satellite during the ejection.

Short answer

a) The speed of the shuttle in the minus x-direction after the ejection is \(0.0028\;{\rm{m/s}}\).

b) The average force exerted on the satellite by the shuttle during the ejection is 53.12 N.

Step-by-step solution

Identification of the given data

The mass of the satellite is\({m_1} = 850\;{\rm{kg}}\).

The time taken by the shuttle to eject the satellite is\(t = 4.8\;{\rm{s}}\).

The launch speed of the satellite is \({v_1} = 0.30\;{\rm{m/s}}\).

The mass of the shuttle is \({m_2} = 92000\;{\rm{kg}}\).

(a) Statement of the principle of conservation of linear momentum

The principle of conservation of linear momentum states that if two bodies collide with each other, the total linear momentum before and after the collision remains the same if no external force acts on the system.

\(\begin{aligned}{c}{p_{{\rm{before}}}} = {p_{{\rm{after}}}}\\{\left( {mv} \right)_{{\rm{before}}}} = {\left( {mv} \right)_{{\rm{after}}}}\end{aligned}\)

(a) Determination of the final speed of the shuttle after ejecting the satellite  

The velocity of the shuttle in the minus x-direction after the ejection of the satellite is\({v_{\rm{f}}}\). Applying the principle of conservation of linear momentum before and after ejecting the satellite, you get:

\(\begin{aligned}{c}{m_1}{v_1} = {m_2}{v_{\rm{f}}}\\{v_{\rm{f}}} = \frac{{{m_1}{v_1}}}{{{m_2}}}\\ = \frac{{\left( {850\;{\rm{kg}}} \right)\left( {0.30\;{\rm{m/s}}} \right)}}{{92000\;{\rm{kg}}}}\\ = 0.0028\;{\rm{m/s}}\end{aligned}\)

Thus, the speed of the shuttle in the minus x-direction is \(0.0028\;{\rm{m/s}}\).

Definition of impulse

Impulse is the change in the momentum of an object.

\(\begin{aligned}{c}J = {\left( {mv} \right)_{{\rm{final}}}} - {\left( {mv} \right)_{{\rm{initial}}}}\\ = m\Delta v\end{aligned}\) … (i)

It is also defined as the product of the force applied on an object and the time interval during which the force is applied.

\(\begin{aligned}{c}J = F{t_{{\rm{final}}}} - F{t_{{\rm{initial}}}}\\ = F\Delta t\end{aligned}\) … (ii)

(b) Determination of the average force exerted on the satellite

Compare equations (i) and (ii).

\(\begin{aligned}{c}m\Delta v = F\Delta t\\F = \frac{{m\Delta v}}{{\Delta t}}\\ = \frac{{\left( {850\;{\rm{kg}}} \right)\left( {0.30\;{\rm{m/s}}} \right)}}{{4.8\;{\rm{s}}}}\\ = 53.12\;{\rm{N}}\end{aligned}\)

Thus, the average force exerted on the satellite is 53.12 N.